Question

prove that the point (-1,2) is the centre of a circle passing through the point (-3,11),(5,9),(8,0) and (6,8). find the radius of the circle​

Answer 1

Answer:

Step-by-step explanation:

Find the distance between the centre and each point. U will see all ar coming equal hence proved as they all are the radius of the circle

Answer 2

Answer:

Center: (-1,2)

Radius = $\sqrt{85}$

Step-by-step explanation:

A circle passing through the points (-3,11),(5,9),(8,0) and (6,8)

Let center of circle be (h,k)

Equation of circle: $(x-h)^2+(y-k)^2=r^2$

using passing point to find r and centre

For (-3,11)

$(-3-h)^2+(11-k)^2=r^2$-------------(1)

For (5,9)

$(5-h)^2+(9-k)^2=r^2$--------------(2)

For (8,0)

$(8-h)^2+(0-k)^2=r^2$--------------(3)

Now solve system of equations (1), (2) and (3) and we get

Subtract eq(1) - eq(2)

$(-3-h)^2-(5-h)^2+(11-k)^2-(9-k)^2=r^2-r^2$

$(2-2h)(-8)+(20-2k)(2)=0$

$4h-k=-6$-----------------------------(4)

Subtract eq(3) - eq(2)

$(8-h)^2-(5-h)^2+(0-k)^2-(9-k)^2=r^2-r^2$

$(13-2h)(3)+(9-2k)(-9)=0$

$h-3k=-7$-----------------------------(5)

Solve eq(4) and eq(5) for h and k

substitute k from equation 4 to eq(5)

$h-3(4h+6)=-7$

$h=-1$

$k=2$

Center: (-1,2)  hence proved

For center put center and passing point in equation (3)

(h,k) ⇔ (-1,2) , Passing point (6,8)

$(6+1)^2+(8-2)^2=r^2$

$r^2=49+36$

$r=\sqrt{85}$

Hence, The center of given circle is (-1,2) and radius is $\sqrt{85}$